Problem: Find the $\emph{positive}$ real number(s) $x$ such that $\frac{1}{2}\left( 3x^2-1\right) = \left( x^2-50x-10\right)\left( x^2+25x+5\right)$.
Explanation: Write $a = x^2-50x-10$ and $b = x^2+25x+5$.  Then the equation given becomes
\[\frac{a+2b-1}{2} = ab,\]so $0=2ab-a-2b+1=(a-1)(2b-1)$. Then $a-1=x^2-50x-11=0$ or $2b-1=2x^2+50x+9=0$. The former has a positive root, $x=\boxed{25 + 2\sqrt{159}}$, while the latter does not.